On the Möbius $\mu$ function A search on wikipedia shows:
$$\mu(n) = \sum_{k=1,gcd(k,n)=1}^{n} e^{2\pi i \frac{k}{n}}$$
But that uses complex numbers... and requires finding out the gcd... 
How useful will be a method, if that could find the exact value for $\mu(n)$ function, by just knowing all the values from $\mu(1)$ till $\mu(n-1)$, which does not require factoring any integer and which uses only elementary methods?
Has this been done before? My question is more precisely:

Does any such formula exist?

 A: (This should be a comment.)
FWIW, you can derive from the series you gave the alternative representation
$$\mu(n)=\sum_{j=1}^n [\gcd(n,k)=1]\cos\left(\frac{2\pi k}{n}\right)$$
where $[p]$ is an Iversonian bracket for the condition $p$.
Since checking if a number is squarefree is conjectured to be at least as hard as straight factorization, I don't have much hope of an algorithm of the sort you want existing. (If there is, I too would be interested in hearing about it. ;) )
A: The Möbius mu function is denoted by $\mu(x)$.
Another function that I got while studying Dirichlet eta function
$$\nu(x) = \rho\left(\frac{x}{2}\right)+\frac{1}{2} - \rho\left(\frac{x}{2}+ \frac{1}{2}\right)$$
where, $\rho(x)$ is the fractional part of $x$. Observe $\nu(x)$ only takes the values $0$ and $1$
While working with a criterion (equivalent to the Nyman-Beurling approach) for Dirichlet eta function, I observed,
$$\sum_{k=1}^{n} \mu(k) \nu(n/k) = -1 $$ for all $n \geq 2$
It has given me a recursive algorithm to calculate $\mu$ since calculating $\nu$ is much easier.
We start with
$$f_1(x) = \nu(x)$$
$$f_n(x) = f_{n-1}(x) + \mu(n)\nu(x/n)$$
Then,
$$\mu(n) = -1 - f_{n-1}(n)$$
